TITLE: Do torsors give a long exact sequence of cohomology?
QUESTION [7 upvotes]: Let $X$ be a finite-type scheme over a field $k$. Let $G$ be a finite-type group scheme over $k$; we write $G_X$ for the base-change of $G$ from $\operatorname{Spec}(k)$ to $X$.
Suppose $f : Y \rightarrow X$ is a $G_X$-torsor for the fppf topology (i.e. we have an $X$-group scheme action of $G_X$ on $Y$ such that the morphism $G_X \times_X Y \rightarrow Y \times_X Y$ given on points by $(g,y) \mapsto (y,gy)$ is an isomorphism). Such a $Y$ gives a class $[Y]$ in the fppf cohomology set $H^1(X,G_X)$ that classifies fppf $G_X$-torsor sheaves over $X$ (this is defined with Cech cohomology).
Consider the specialization map
$$
s : X(k) \rightarrow H^1(k,G)
$$
that sends $x : \operatorname{Spec}(k) \rightarrow X$ to the pull-back of $[Y]$ by $x$, which is an element of $H^1(k,G)$. Note: if $G$ is smooth, the fppf cohomology set $H^1(k,G)$ may be identified with the Galois cohomology set $H^1(k,G(k^{\mathrm{sep}}))$.
Supposing it exists, fix a $y \in Y(k)$. We obtain an exact sequence of pointed sets
$$
0 \rightarrow G(k) \rightarrow Y(k) \stackrel{f}{\rightarrow} X(k) \stackrel{s}{\rightarrow} H^1(k,G)
$$
where $G(k) \rightarrow Y(k)$ is just the inclusion of the fiber above $y$. From the looks of it, I'd say that this has to be the start of a long exact sequence of some kind. The question only is: what kind? I don't see an obvious way of continuing it, since $Y$ doesn't necessarily carry any group structure so as to give meaning to the expression $H^1(k,Y)$.
Question: Is this exact sequence part of a long exact sequence? For instance, are we witnessing some instantiation of homotopy theory? If not, is there any other more conceptual way of viewing the above sequence?
Example: As a motivating example, let me show you why the image of $s$ - and therefore a continuation of the exact sequence from above - is an interesting object of study. Let $E$ be an elliptic curve over $\mathbf{Q}$ given by $y^2=f(x)$. Let $Y$ be $E - E[2]$, let $X$ be $\operatorname{Spec}(\mathbf{Q}[x,f^{-1}])$ (i.e., the affine line with coordinate $x$ and with the subscheme $f=0$ deleted), and let $f:Y \rightarrow X$ be the map that sends $(x,y)$ to $x$. Let $G = \mu_2$. We endow $Y$ with the structure of a $G$-torsor by letting the non-trivial element of $G$ send $(x,y)$ to $(x,-y)$. Then the image of the map
\begin{align*}
X(\mathbf{Q}) & \rightarrow \mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2} ~~ (\cong H^1(\mathbf{Q},\mu_2)) \\\
x & \mapsto f(x) \pmod{\mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2}}
\end{align*}
consists precisely of those elements $c \in \mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2}$ such that $cy^2=f(x)$ contains rational points other than the "trivial ones", i.e. the zeros of $f$ and the point at infinity.

REPLY [8 votes]: It actually more like the ending of a long exact sequence, rather than the beginning. To see what's going on consider the analogous case in topology. For this you replace the Galois group of $k$ with a discrete group $\Gamma$ and the category of $k$-schemes with the category of $\Gamma$-spaces. Instead of an algebraic group you now have a topological group G equipped with an action of $\Gamma$ on its classifying space BG. A G-torsor is a principle fibration $Y \to X$, which in homotopy theory corresponds to a $\Gamma$-equivariant fibration sequence of the form
$$ Y \to X \to BG $$
Taking $\Gamma$-homotopy fixed points one obtains a fibration sequence
$$ Y^{h\Gamma} \to X^{h\Gamma} \to BG^{h\Gamma} $$
which leads to a long exact sequence of homotopy groups ending with
$$ ... \to \pi_1(BG^{h\Gamma}) \to \pi_0(Y^{h\Gamma}) \to \pi_0(X^{h\Gamma}) \to \pi_0(BG^{h\Gamma}) $$ 
where $\pi_1(BG^{h\Gamma}) = \pi_0(G^{h\Gamma})$. This tail is the homotopy theoretic analogue of the sequence
$$ G(k) \to Y(k) \to X(k) \to H^1(k,G) $$
and so one should consider this sequence as the ending, and not the beginning of a long exact sequence. However, as apposed to the homotopy theoretic analogue, the map $G(k) \to Y(k)$ is always injective, making it seems like the sequence is just starting.
The analogy with the homotopy theoretic case can be made more precise by considering the etale homotopy type (see http://arxiv.org/abs/1110.0164).